Let $F$ be a field and $K/F$ be a field extension. Let $\alpha_1, \alpha_2, \dots, \alpha_k$ be algebraic elements of $K$ over $F$.
I am wondering if we have the equality:
\begin{equation*} F(\alpha_1, \alpha_2, \dots ,\alpha_k) = \left\{\sum_{i=1}^m b_i \alpha_1^{n_{1_i}} \cdots \alpha_k^{n_{k_i}} : b_i \in F \text{ and } m \in \mathbb{N} \text{ and } n_j \in \mathbb{Z}\right\} \end{equation*}
For notational ease, let $L$ denote the set on the right hand side.
[Note: The definition of the field $F(\alpha_1, \cdots, \alpha_k)$ that I am working with is that it is the smallest subfield of $K$ containing $F$ and the $\alpha_j$.]
The direction $(\supseteq)$ is fairly obvious, since the set on the left is a field. I have attempted to prove the other direction as follows:
The set $L$ is a subset of $K$ that contains $F$ and the $\alpha_i$. So if we can show that $L$ is a field, we will have the $(\subseteq)$ direction.
The set $L$ is clearly a subgroup of $K$.
Moreover, it is not too hard to see that the set $L$ is closed under multiplication.
We can also see that all of the $\alpha_j$ contain a multiplicative inverse in $L$. Below is a proof for $\alpha_1$, and the proof for the other $\alpha_j$ is similar.
We know that $\alpha_1$ is algebraic over $F$, so there exists some polynomial $m_{\alpha_1}(x) \in F[x]$ with $m_{\alpha_1}(\alpha_1) = 0$. Let $m_{\alpha_1}(x) = b_rx^r + b_{r-1}x^{r-1} + \cdots + b_1 x + b_0$. Then we get
\begin{align*} 0 & = m_{\alpha_1}(\alpha_1) \\ & = b_r(\alpha_1)^r + \cdots + b_1 \alpha_1 + b_0 \\ & = \alpha_1(b_r \alpha_1^{r-1} + b_{r-1} \alpha_1^{r-2}+ \cdots + b_1) + b_0 \end{align*} which gives \begin{equation*} 1 = (\alpha_1)(-b_0^{-1})\Big(b_r \alpha_1^{r-1} + b_{r-1}\alpha_1^{r-2} + \cdots + \alpha_1\Big) \end{equation*} so that $\alpha_1$ does indeed have an inverse in $L$.
However, I am at a complete loss of how to show that an arbitrary element of $L$ contains a multiplicative inverse in $L$.
Just to simplify typing I use only two algebraic elements $a, b$ in $K$ over $F$ and show that $F(a, b)$ has the desired form. Note that $F(a, b) = F(a)(b)$ and clearly each element of $F(a)$ has the form $$\beta_{0} + \beta_{1}a + \beta_{2}a^{2} + \cdots + \beta_{n - 1}a^{n - 1}$$ where $n$ is degree of minimal polynomial of $a$ over $F$. Now $b$ is algebraic over $F$ and hence over $F(a)$ also and let $m$ be the degree of minimal polynomial of $b$ over $F(a)$. Then elements of $F(a)(b)$ look like $$c_{0} + c_{1}b + \cdots + c_{m - 1}b^{m - 1}$$ where $c$'s belong to $F(a)$ and are hence expressible as polynomials in $a$. Thus the elements of $F(a, b)$ are expressible as polynomials in $a, b$ with coefficients in $F$. This means that you can keep all $n_j$ as non-negative.